This looks hopeless to have a nice formula isn't it ?
–
camomilleMar 20 '11 at 14:39

2

You definitely must tell more about the graph (and about the weight function) to get any answer at all. Compare the case of the line graph with $v_1$ and $v_2$ far apart to the case of the complete graph on $n$ vertices with $n$ large.
–
DidMar 20 '11 at 14:52

@Didier well, you right, that's implied part of the question -- if there are no nice results for arbitrary graph, maybe there are any non-trivial classes of graphs, where this problem is trackable? In my context it would be randomly generated scale-free network with number of edges that makes the brute force method unfeasible.
–
alystMar 20 '11 at 17:33

Thanks, Robert. That gives an idea for an alternative formula: if $M$ is a collection of all minimal sets of edges, which removal disrupts connectivity between $v_1$ and $v_2$, then $1 - \sum_{B \subseteq M} (-1)^{|B|-1} P(\cup B \not\subseteq E')$ should also be the sought probability, where $P(S \not\subseteq E') = \prod_{e \in S} (1-p(e))$.
–
alystMar 20 '11 at 21:28